I am working on the following integral
$\int_0^1 dx\int_0^1 dT \sqrt{1-(1-\sqrt{x}+\sqrt{xT})^2} e^{-n xT},$
as $n\rightarrow \infty$. The goal is to find the asymptotic behavior of the integral to the leading order of $n$.
Obviously, there is a saddle point at $(x,T)=(0,0)$, which is the main difficulty of this calculation. It seems this is related to the Laplace's method introduced in particular in Chapter VIII of [Wong, R. (2001). Asymptotic approximations of integral]. But when I try to change the variable $y_1=(x+T)/2$, $y_2=(x-T)/2$, $y_1=\sqrt{\xi}\cosh{\mu}$, $y_2=\sqrt{\xi}\cosh{\mu}$. It seems all the higher order terms of $\xi$ in the expansion of $\sqrt{1-(1-\sqrt{x}+\sqrt{xT})^2}$ will contribute to the results.
Thanks for the satisfying solutions. I think the question is already addressed. This question is made up by myself with the goal of addressing a more complex problem posted here
Follow up question about Asymptotic behavior of integral with Laplace's method
This may need some more efforts and possibly a more general method.
Best Answer
Elaborating on Maxim's comment. If $v=xT$, then \begin{align*} I(n)&=\int_0^1\!\! {\int_0^1 {\mathrm{e}^{ - nxT} \sqrt {1 - (1 - \sqrt x + \sqrt {xT} )^2 } \,\mathrm{d}T}\, \mathrm{d}x} \\& = \int_0^1\!\! {\int_0^x {\mathrm{e}^{ - nv} \frac{{\sqrt {1 - (1 - \sqrt x + \sqrt v )^2 } }}{x}\,\mathrm{d}v} \,\mathrm{d}x} \\ & = \int_0^1 {\mathrm{e}^{ - nv} \int_v^1 {\frac{{\sqrt {1 - (1 - \sqrt x + \sqrt v )^2 } }}{x}\,\mathrm{d}x} \,\mathrm{d}v} . \end{align*} Now, \begin{align*} & \int_v^1 {\frac{{\sqrt {1 - (1 - \sqrt x + \sqrt v )^2 } }}{x}\,\mathrm{d}x} \\ & = 2(\sqrt {1 - v} + (1 + \sqrt v )\arccos (\sqrt v ) - v^{1/4} \sqrt {2 + \sqrt v } \arccos (v + \sqrt v - 1)) \\ & = (\pi + 2) - 2\sqrt 2 \pi v^{1/4} + (\pi + 2)v^{1/2} - \frac{{\sqrt 2 }}{2}\pi v^{3/4} + \frac{1}{3}v + \ldots \end{align*} as $v\to 0^+$. Thus, by Watson's lemma, $$ I(n) \sim \frac{{\pi + 2}}{n} - \frac{{\sqrt 2 \pi \Gamma (1/4)}}{{2n^{5/4} }} + \frac{{(\pi + 2)\sqrt \pi }}{{2n^{3/2} }} - \frac{{3\sqrt 2 \pi \Gamma (3/4)}}{{8n^{7/4} }} + \frac{1}{{3n^2 }} + \ldots $$ as $n\to +\infty$.